Question

Problem 10-2. (fig. 10-7): The length of the top view
of a line parallel to the V.P. and inclined at 45 to the H.P.
is 50 mm. One end of the line is 12 mm above the H.P.
and 25 mm in front of the VP Draw the projections of
the line and determine its true length.​

Answer 1

True length of a'b' is 75mm

Answer 2

Answer:

The true length of the line segment is sqrt(1949) mm.

Explanation:

First, we draw the top view of the line, which is a line segment inclined at 45 degrees to the horizontal plane (H.P.) and parallel to the vertical plane (V.P.). We know that the length of this line segment in the top view is 50 mm and one end of the line is 12 mm above the H.P. and 25 mm in front of the V.P. We can represent the line segment in the top view as AB, where A is the end of the line segment that is 25 mm in front of the V.P., and B is the other end of the line segment.

To draw the projections of the line, we first draw the front view of the line. Since the line is parallel to the V.P., its front view is a point on the V.P. directly below the top view. Let C be the point on the V.P. that corresponds to the line in the front view.

Next, we draw the side view of the line. Since the line is inclined at 45 degrees to the H.P., its side view is a line that makes an angle of 45 degrees with the H.P. and passes through the end point A in the top view. Let D be the point where the line intersects the V.P. in the side view.

To determine the true length of the line segment, we need to find the distance between the end points A and B in the space. To do this, we draw a line from A to D in the side view, and project it down to the H.P. in the top view. Let E be the point where the projected line intersects the H.P.

We can now use the Pythagorean theorem to find the true length of the line segment. We have:

AB^2 = AE^2 + EB^2

We know that AE = 12 mm, since A is 12 mm above the H.P. We also know that EB = 50 mm, since that is the length of the line segment in the top view. To find AE, we need to find the length of the projected line AD in the side view. Since AD makes an angle of 45 degrees with the H.P., we have:

AD = AB/sqrt(2) = 50/sqrt(2) mm

Therefore, we have:

AE = AD*cos(45) = (50/sqrt(2))*cos(45) = 25 mm

Substituting these values into the equation for AB^2, we get:

AB^2 = 12^2 + 25^2 + 50^2

AB = sqrt(12^2 + 25^2 + 50^2) = sqrt(1949) mm

Therefore, the true length of the line segment is sqrt(1949) mm.

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